1. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 3 Polynomial and Rational Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2. 2 3.4 Polynomial Functions: Graphs, Applications, and Models Graphs of  (x) = ax n Graphs of General Polynomial Functions Behavior at Zeros Turning Points and End Behavior Graphing Techniques Intermediate Value and Boundedness Theorems Approximating Real Zeros Polynomial Models

3. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 Graphs of General Polynomial Functions As with quadratic functions, the absolute value of a in  (x) = ax n determines the width of the graph. When  a  > 1, the graph is stretched vertically, making it narrower. When 0 <  a  < 1, the graph is shrunk or compressed vertically, making it wider.

4. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 Graphs of General Polynomial Functions Compared with the graph of f (x) = ax n, the following also hold true. The graph of  (x) = – ax n is reflected across the x-axis The graph of f (x) = ax n + k is translated (shifted) k units up if k > 0 and  k  units down if k < 0. The graph of  (x) = a(x – h) n is translated h units to the right if h > 0 and  h  units to the left if h < 0. The graph of  (x) = a(x – h) n + k shows a combination of these translations.

5. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 Example 1 EXAMINING VERTICAL AND HORIZONTAL TRANSLATIONS Solution The graph will be the same as that of  (x) = x 5, but translated 2 units down. This function is increasing on its entire domain (–∞,∞). (a) Graph each function. Determine the intervals of the domain for which each function is increasing or decreasing.

6. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 Example 1 Solution In  (x) = (x + 1) 6, function  has a graph like that of  (x) = x 6, but since x + 1 = x – (– 1), it is translated 1 unit to the left. This function is decreasing on (–∞,–1] and increasing on [–1,∞). (b) EXAMINING VERTICAL AND HORIZONTAL TRANSLATIONS Graph each function. Determine the intervals of the domain for which each function is increasing or decreasing.

7. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 Example 1 Solution The negative sign in – 2 causes the graph of the function to be reflected across the x-axis when compared with the graph of  (x) = x 3. Because  – 2  > 1, the graph is stretched vertically when compared to the graph of  (x) = x 3. It is also translated 1 unit to the right and 3 units up. This function is increasing on its entire domain (–∞,∞). (c) EXAMINING VERTICAL AND HORIZONTAL TRANSLATIONS Graph each function. Determine the intervals of the domain for which each function is increasing or decreasing.

8. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 Unless otherwise restricted, the domain of a polynomial function is the set of all real numbers. Polynomial functions are smooth, continuous curves on the interval (– ,  ). The range of a polynomial function of odd degree is also the set of all real numbers. Typical graphs of polynomial functions of odd degree are shown next. These graphs suggest that for every polynomial function  of odd degree there is at least one real value of x that makes  (x) = 0. The real zeros are the x-intercepts of the graph and can be determined by inspecting the factored form of each polynomial.

9. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 Odd Degree

10. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 A polynomial function of even degree has a range of the form (– , k] or [k,  ) for some real number k. Here are two typical graphs. Even Degree

11. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 The behavior of a graph at a zero depends on the multiplicity of the zero as determined by the exponent on the corresponding factor. The graph crosses the x-axis at (c, 0) if c is a zero of multiplicity 1.

12. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 The graph is tangent to the x-axis at (c, 0) if c is a zero of even multiplicity. The graph bounces, or turns, at c.

13. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 The graph crosses and is tangent to the x-axis at (c, 0) if c is a zero of odd multiplicity greater than 1. The graph wiggles at c.

14. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Turning Points and End Behavior The graphs exhibiting functions of odd degree and even degree show that polynomial functions often have turning points where the function changes from increasing to decreasing or from decreasing to increasing.

15. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Turning Points A polynomial function of degree n has at most n – 1 turning points, with at least one turning point between each pair of successive zeros.

16. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 End Behavior The end behavior of a polynomial graph is determined by the dominating term—that is, the term of greatest degree. A polynomial of the form has the same end behavior as.

17. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 End Behavior For example, has the same end behavior as. It is large and positive for large positive values of x and large and negative for negative values of x with large absolute value. That is, it rises to the right and falls to the left.

18. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 End Behavior This figure shows that as x increases without bound, y does also. As and as

19. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 End Behavior The graph in this figure has the same end behavior as f(x) = –x 3. As and as

20. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 End Behavior of Graphs of Polynomials Suppose that ax n is the dominating term of a polynomial function  of odd degree. 1. If a > 0, then as and as Therefore, the end behavior of the graph is of the type shown here. We symbolize it as.

21. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 End Behavior of Polynomials Suppose that ax n is the dominating term of a polynomial function  of odd degree. 2. If a < 0, then as and as Therefore, the end behavior of the graph is of the type shown here. We symbolize it as.

22. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 End Behavior of Polynomials Suppose that ax n is the dominating term of a polynomial function  of even degree. 1. If a > 0, then as Therefore, the end behavior of the graph is of the type shown here. We symbolize it as.

23. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 End Behavior of Polynomials Suppose that is the dominating term of a polynomial function  of even degree. 2. If a < 0, then as Therefore, the end behavior of the graph is of the type shown here. We symbolize it as.

24. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 2 DETERMINING END BEHAVIOR Based on the previous discussion of end behavior, match each function with its graph. Solution Because  is of even degree with positive leading coefficient, its graph is in C. A.B.C. D.

25. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 2 Solution Because g is of even degree with negative leading coefficient, its graph is in A. A.B.C. D. DETERMINING END BEHAVIOR Based on the previous discussion of end behavior, match each function with its graph.

26. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Example 2 Solution Because function h has odd degree and has a dominating term positive coefficient, its graph is in B. A.B.C. D. DETERMINING END BEHAVIOR Based on the previous discussion of end behavior, match each function with its graph.

27. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 2 Solution Because function k has odd degree and a dominating term with negative coefficient, its graph is in D. A.B.C. D. DETERMINING END BEHAVIOR Based on the previous discussion of end behavior, match each function with its graph.

28. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Graphing Techniques We have discussed several characteristics of the graphs of polynomial functions that are useful for graphing the function by hand. A comprehensive graph of a polynomial function will show the following characteristics. 1. all x-intercepts (zeros) and the behavior of the graph at these zeros 2. the y-intercept 3. the sign of  (x) within the intervals formed by the x-intercepts 4. enough of the domain to show the end behavior In Examples 3 and 4, we sketch the graphs of two polynomial functions by hand. While there are several ways to approach this process, we use the following general guidelines.

29. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Graphing a Polynomial Function Let be a polynomial function of degree n. To sketch its graph, follow these steps. Step 1 Find the real zeros of . Plot them as x-intercepts. Step 2 Find  (0) = a 0. Plot this as the y-intercept. Step 3 Use end behavior, whether the graph crosses, bounces on, or wiggles through the x-axis at the x-intercepts, and selected points as necessary to complete the graph.

30. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 1 The possible rational zeros are  1,  2,  3,  6,  1/2, and  3/2. Use synthetic division to show that 1 is a zero.  (1) = 0

31. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 1 Thus, Factor. Set each linear factor equal to 0, then solve for x to find zeros. The three zeros of  are 1, – 3/2, and – 2.

32. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 2

33. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 Since the dominating term of  (x) is 2x 3, the graph will have end behavior similar to that of f (x) = x 3. It will rise to the right and fall to the left as. Each zero of f (x) occurs with multiplicity 1, meaning that the graph of f (x) will cross the x-axis at each of its zeros. Because the graph of a polynomial function has no breaks, gaps, or sudden jumps, we now have sufficient information to sketch the graph of f (x).

34. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 Begin sketching at either end of the graph with the appropriate end behavior, and draw a smooth curve that crosses the x-axis at each zero, has a turning point between successive zeros, and passes through the y-intercept as shown.

35. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 Additional points may be used to verify whether the graph is above or below the x-axis between the zeros and to add detail to the sketch of the graph. The zeros divide the x-axis into four intervals: (– , –2), (–2, –3/2), (–3/2, 1), and (1,  ). Select an x-value as a test point in each interval, and substitute it into the equation for f (x) to determine additional points on the graph.

36. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 A typical selection of test points and the results of the tests are shown in the table.

37. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Example 3 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 The sketch could be improved by plotting the points found in each interval in the table.

38. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 4 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 1 Since the polynomial is given in factored form, the zeros can be determined by inspection. They are 1, 3, and –2. Plot these as x-intercepts of the graph of f(x).

39. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 4 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 2 Plot (0,–12) on the y-axis.

40. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 40 Example 4 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 The dominating term of f (x) can be found by multiplying the factors and identifying the term of greatest degree. Here it is –(x)(x)(x) 2 = –x 4, indicating that the end behavior of the graph is. Because 1 and 3 are zeros of multiplicity 1, the graph will cross the x-axis at these zeros. The graph of f (x) will touch the x-axis at –2 and then turn and change direction because it is a zero of even multiplicity.

41. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 41 Example 4 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 Begin at either end of the graph with the appropriate end behavior and draw a smooth curve that crosses the x-axis at 1 and 3 and that touches the x-axis at –2, then turns and changes direction. The graph will also pass through the y-intercept –12.

42. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 42 Example 4 GRAPHING A POLYNOMIAL FUNCTION Graph Solution Step 3 Using test points within intervals formed by the x-intercepts is a good way to add detail to the graph and verify the accuracy of the sketch. A typical selection of test points is (–3,–24), (–1,–8), (2, 16), and (4,–108).

43. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 43 Graphing Polynomial Functions Note It is possible to reverse the process of Example 4 and write the polynomial function from its graph if the zeros and any other point on the graph are known. Suppose that you are asked to find a polynomial function of least possible degree having the graph shown.

44. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 44 Graphing Polynomial Functions Note Because the graph crosses the x-axis at 1 and 3 and bounces at –2, we know that the factored form of the function is as follows. Multiplicity one Multiplicity two

45. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 45 Graphing Polynomial Functions Note Now find the value of a by substituting the x- and y-values of any other point on the graph, say (0,–12), into this function and solving for a. Verify in Example 4 that the polynomial function is f(x) = –(x – 1)(x – 3)(x + 2) 2. Let x = 0 and y = –12. Simplify. Divide by 12.

46. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 46 Important Relationships We emphasize the important relationships among the following concepts. the x-intercepts of the graph of y =  (x) the zeros of the function  the solutions of the equation  (x) = 0 the factors of  (x)

47. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 47 x-Intercepts, Zeros, Solutions, and Factors If f is a polynomial function and a is an x-intercept of the graph of then a is a zero of , a is a solution of  (x) = 0, and x – a is a factor of  (x).

48. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 48 Intermediate Value Theorem for Polynomials If  (x) defines a polynomial function with only real coefficients, and if for real numbers a and b, the values  (a) and  (b) are opposite in sign, then there exists at least one real zero between a and b.

49. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 49 Example 5 LOCATING A ZERO Use synthetic division and a graph to show that  (x) = x 3 – 2x 2 – x + 1 has a real zero between 2 and 3. Solution Since  (2) is negative and  (3) is positive, by the intermediate value theorem there must be a real zero between 2 and 3.

50. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 50 Example 5 LOCATING A ZERO Use synthetic division and a graph to show that  (x) = x 3 – 2x 2 – x + 1 has a real zero between 2 and 3. Solution The graphing calculator screen indicates that this zero is approximately 2.2469796. (Notice that there are two other zeros as well.)

51. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 51 Caution Be careful how you interpret the intermediate value theorem. If  (a) and  (b) are not opposite in sign, it does not necessarily mean that there is no zero between a and b. In the graph shown here, for example,  (a) and  (b) are both negative, but – 3 and – 1, which are between a and b, are zeros of  (x).

52. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 52 Boundedness Theorem Let  (x) be a polynomial function of degree n ≥ 1 with real coefficients and with a positive leading coefficient. Suppose  (x) is divided synthetically by x – c. (a) If c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then  (x) has no zero greater than c.

53. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 53 Boundedness Theorem (b) if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then  (x) has no zero less than c.

54. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 54 Proof of the Boundedness Theorem We outline the proof of part (a). The proof for part (b) is similar. By the division algorithm, if  (x) is divided by x – c, then for some q (x) and r, where all coefficients of q (x) are nonnegative, r ≥ 0, and c > 0. If x > c, then x – c > 0. Since q (x) > 0 and r ≥ 0, This means that  (x) will never be 0 for x > c.

55. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 55 Example 6 USING THE BOUNDEDNESS THEOREM Solution Since  (x) has real coefficients and the leading coefficient, 2, is positive, use the boundedness theorem. Divide  (x) synthetically by x – 3. (a) No real zero is greater than 3. Show that the real zeros of  (x) = 2x 4 – 5x 3 + 3x + 1 satisfy these conditions. All are nonnegative.

56. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 56 Example 6 USING THE BOUNDEDNESS THEOREM Solution (a) No real zero is greater than 3. All are nonnegative. Since 3 > 0 and all numbers in the last row of the synthetic division are nonnegative,  (x) has no real zero greater than 3. Show that the real zeros of  (x) = 2x 4 – 5x 3 + 3x + 1 satisfy these conditions.

57. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 57 Example 6 USING THE BOUNDEDNESS THEOREM Solution (b) No real zero is less than –1. These numbers alternate in sign. Divide  (x) by x + 1. Here –1 < 0 and the numbers in the last row alternate in sign, so  (x) has no real zero less than –1. Show that the real zeros of  (x) = 2x 4 – 5x 3 + 3x + 1 satisfy these conditions.

58. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 58 Example 7 APPROXIMATING REAL ZEROS OF A POLYNOMIAL FUNCTION Approximate the real zeros of  (x) = x 4 – 6x 3 + 8x 2 + 2x – 1. Solution The dominating term is x 4, so the graph will have end behavior similar to the graph of  (x) = x 4, which is positive for all values of x with large absolute values. That is, the end behavior is up at the left and the right,. There are at most four real zeros, since the polynomial is fourth-degree.

59. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 59 Example 7 APPROXIMATING REAL ZEROS OF A POLYNOMIAL FUNCTION Approximate the real zeros of  (x) = x 4 – 6x 3 + 8x 2 + 2x – 1. Solution Since  (0) = – 1, the y-intercept is –1. Because the end behavior is positive on the left and the right, by the intermediate value theorem  has at least one zero on either side of x = 0. To approximate the zeros, we use a graphing calculator.

60. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 60 Example 7 APPROXIMATING REAL ZEROS OF A POLYNOMIAL FUNCTION Approximate the real zeros of  (x) = x 4 – 6x 3 + 8x 2 + 2x – 1. Solution The graph below shows that there are four real zeros, and the table indicates that they are between – 1 and 0, 0 and 1, 2 and 3, and 3 and 4 because there is a sign change in  (x) in each case.

61. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 61 Example 7 APPROXIMATING REAL ZEROS OF A POLYNOMIAL FUNCTION Solution Using the capability of the calculator, we can find the zeros to a great degree of accuracy. The graph shown here shows that the negative zero is approximately – 0.4142136. Similarly, we find that the other three zeros are approximately 0.26794919, 2.4142136, and 3.7320508.

62. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 62 Example 8 EXAMINING A POLYNOMIAL MODEL The table shows the number of transactions, in millions, by users of bank debit cards for selected years.

63. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 63 Example 8 EXAMINING A POLYNOMIAL MODEL (a) Using x = 0 to represent 1995, x = 3 to represent 1998, and so on, use the regression feature of a calculator to determine the quadratic function that best fits the data. Plot the data and the graph.

64. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 64 Example 8 EXAMINING A POLYNOMIAL MODEL (a) The best-fitting quadratic function for the data is defined by Solution

65. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 65 Example 8 EXAMINING A POLYNOMIAL MODEL (b) Repeat part (a) for a cubic function (degree 3). Solution The best-fitting cubic function is shown and is

66. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 66 Example 8 EXAMINING A POLYNOMIAL MODEL (c) Repeat part (a) for a quartic function (degree 4). The best-fitting quartic function is shown and is Solution

67. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 67 Example 8 EXAMINING A POLYNOMIAL MODEL (d) The correlation coefficient, R, is a measure of the strength of the relationship between two variables. The values of R and R 2 are used to determine how well a regression model fits a set of data. The closer the value of R 2 is to 1, the better the fit. Compare R 2 for the three functions to decide which function best fits the data.

68. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 68 Example 8 EXAMINING A POLYNOMIAL MODEL With the statistical diagnostics turned on, the value of R 2 is displayed with the regression results on the TI 83/84 Plus each time that a regression model is executed. By inspecting the R 2 value for each model above, we see that the quartic function provides the best fit since it has the largest R 2 value of 0.9999026772. Solution

69. Copyright © 2013, 2009, 2005 Pearson Education, Inc. 69 Note In Example 8(d), we selected the quartic function as the best model based on the comparison of R 2 values of the models. In practice, however, the best choice of a model should also depend on the set of data being analyzed as well as analysis of its trends and attributes.